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Is it possible to calculate the following definite integral in a closed form?

$$ \int_0^\infty \left| \sin x \cdot \sin (\pi x) \right| e^{-x} \, dx$$

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Wolfram Alpha doesn't seem to be able to do it... –  Daniel Freedman Dec 14 '11 at 0:46
You could evaluate the indefinite integral, compute the definite integral over each "period" and then write as a sum. Not sure if this gets you anywhere. –  dls Dec 14 '11 at 1:15
The portion of the integrands with the sines is not periodic, so the prognosis is not great. –  ncmathsadist Dec 14 '11 at 2:47
Maple gives the result $\frac{2\pi}{4+\pi^4}$. –  Jon Dec 14 '11 at 8:32
@Jon: that's the result if the integral didn't have absolute value signs in it. Numerically the OP's integral is a little less than 0.359, while $2\pi/(4+\pi^4) < 0.1$. –  Greg Martin Dec 14 '11 at 8:40

2 Answers 2

One could give the following a try: Develop $|\sin x|$ into a Fourier series. You get $$|\sin x|={2\over\pi}-{4\over\pi}\sum_{k=1}^\infty {1\over 4k^2 -1}\cos(2kx)\ .$$ Similarly $$|\sin (\pi x)|={2\over\pi}-{4\over\pi}\sum_{k=1}^\infty {1\over 4k^2 -1}\cos(2\pi kx)\ .$$ Since the two series are absolutely convergent you can multiply them, obtaining a double series of the form $$\sum_{k,l} 2c_{k,l}\cos(2kx)\cos(2\pi l x)=\sum_{k,l} c_{k,l}\bigl(\cos \bigl((2(k+\pi l)x\bigr)+\cos\bigl(2(k-\pi l) x\bigr)\bigr)\ .$$ Now $$\int_0^\infty \cos(q x)e^{-x}\ dx={1\over 1+q^2}\ ;$$ therefore you will end up with a huge double series containing terms of the form $${c\over (4k^2-1)(4l^2-1)\bigl(1+4(k\pm \pi l)^2\bigr)}\ .$$ I wish you luck$\ldots$

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"I wish you luck..." LOL –  Pedro Tamaroff Apr 16 '12 at 0:57

Let $f(x) = e^{-x} \sin(x)\sin(\pi x)$

Let $A=\{x : e^{-x}sin(x)\sin(\pi x) > 0\}$
Let $B=\{x : e^{-x}sin(x)\sin(\pi x) < 0\}$

A and B are disjoint and hence $\int_{0}^{\infty}f(x)=\int_A f \,du + \int_B f\,du$

Range of $f(x)=0$ to $\kappa =\max(f(x))$

Split the range of $f(x)$ into n intervals, $n\rightarrow \infty$ such that

$\displaystyle \int_A f \,du = \lim_{n\to\infty} \sum_{j=1}^{n} \left (\left(j+1\right)\frac{\kappa }{n}-j\frac{\kappa }{n} \right ) \int I_{A_j}$

$\displaystyle \int_B f \,du = \lim_{n\to\infty} \sum_{j=1}^{n} \left (\left(j+1\right)\frac{\kappa }{n}-j\frac{\kappa }{n} \right ) \int I_{B_j}$

$\displaystyle \int_{A+B} f \,du = \lim_{n\to\infty} \frac{\kappa }{n} \sum_{j=1}^{n} \int I_{A_j} + I_{B_j}$

$\displaystyle A_j =\left (\frac{j\kappa }{n} < f(x) < \frac{(j+1)\kappa }{n} \right )$
$\displaystyle B_j =\left (\frac{j\kappa }{n} < -f(x) < \frac{(j+1)\kappa }{n} \right )$

$\displaystyle h(a,x,b) = \begin{cases} 1 &\text{if } |a| < |x| < |b|, \\ 0 &\text{if } otherwise. \end{cases} $

$\displaystyle I_{A_j} =\frac{1}{2}h\left(j\frac{\kappa }{n},f(x),\left(j+1\right)\frac{\kappa }{n} \right ) \left(1 + \frac{\left|f(x)\right|}{f(x)}\right)$

$\displaystyle I_{B_j} =\frac{1}{2}h\left(-1\left(j+1\right)\frac{\kappa }{n},-f(x),-j\frac{\kappa }{n} \right ) \left(1 - \frac{\left|f(x)\right|}{f(x)}\right)$

working on it.

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Could you fill in a few of the details? As a spectator I'm interested in seeing a complete solution, not necessarily doing it myself. –  Antonio Vargas Apr 14 '12 at 23:29
Not sure what this serves; OP is asking for a closed form and not mere existence... –  J. M. Apr 15 '12 at 16:16

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